jmc

To find the sum, we look at the first few powers of 2 modulo 7: $$\begin{array}{rl}{2}^0& \equiv 1,\\ 2^1& \equiv 2,\\ 2^2& \equiv 4,\\ 2^3& \equiv 8\equiv 1(\mathrm{mod}7)\end{array}$$Since $2^3\equiv 1(\mathrm{mod}7)$, the powers of 2 modulo 7 repeat in cycles of 3. Therefore, $$\begin{array}{rl}& 1+2+2^2+2^3+\cdots +2^{100}\\ & \equiv 1+2+4+1+2+4+\cdots +1+2+4+1+2\\ & \equiv (1+2+4)+(1+2+4)+\cdots +(1+2+4)+1+2\\ & \equiv \boxed{3}(\mathrm{mod}7).\end{array}$$